2. A 2.00A current ows through a circular conductor, which has a radius 12.0cm and lies in the x-y plane. When viewed from the +z-axis, the current is owing clockwise. This loop is in the presence of a uniform magnetic eld given by: 3 B=B i j + 2ko

[ Exam W ] [ Exam X ] [ Exam Y ]

3. A point charge Q lies in the x-y plane. Consider an imaginary block (all vertices are right angles) with dimensions 2L x 2L x L, positioned with one of its square faces at on the plane, centered at the point charge. Use Gausss Law to compute the electric ux through the 2L x 2L surface that is not in the x-y plane. [Hint: The charge is not enclosed by the imaginary block surface indicated.] z 2L 2L L x y

If we include a second box of the same dimensions below the x-y plane, then the charge would be enclosed by a cube, with all sides length 2L. Because of the symmetry, the ux through each face of the cube is the same as through every other, so 1/6 of the total ux through the cube is through each face. The total ux is given by Gausss Law, so that gives us the ux through the top face of the block: Q top face = total = 6 6 o

Physics 9C

Midterm #2 Solutions

March 5, 2013

***BONUS COVERAGE!*** The problem states that you should solve it using Gausss Law, but I cant bring myself to penalize you if you are able to do the integral (but as you will see below, you would have to be a whiz at trig substitution to get the last integral). In case you tried to do this (and for your enhanced educational experience!), I present the integral solution for you here. Start with the electric eld. It is a Coulomb eld at the origin, so: + y Q r Q xi Q j + zk = E ( x, y, z ) = r = 3/2 2 3 4 o r 4 o r 4 o ( x 2 + y2 + z 2 ) At the surface in question, z=L. The area vector points in the +z direction, so the dot product only involves the z-component of the electric eld, which is (at the surface): QL 1 Ez ( x, y, L ) = 3/2 2 2 4 o ( x + y + L2 ) The area element is simply dA=dxdy, so we can plug into the ux integral: +L +L QL + L + L dxdy = E d A = Ez dA = Ez dx dy = 3/2 4 o L L ( x 2 + y 2 + L2 ) L L We do each integral in turn. First we do dx, keeping y constant. This integral happened to be on the integration sheet, so we can use it (shown below) to do the integral over dx to get: dx 1 x a 2 y 2 + L2 , = 2 + const 3/2 2 2 2 2 a x + a x + a ( )

QL + L 1 2L = dy y 2 + L2 y 2 + 2 L2 4 o L We are not so lucky with this next integral it was not on the equation sheet (have I mentioned this was not the intended method?). The integral is symmetric over the -L to +L interval, so we can just make this integral from 0 to L and double it. Making another couple of substitutions puts the integral into a form we can look up, so after all the math we get our answer: du 2 2 u y 2 + L2 , dy = , y = [ 0, L ] u = L ,2 L 2 2 uL

QL 2 L = 2 4 o L 2 QL2 = 2 o bL :2 2 L2

du 2 uL 1 2

1 u 12 4

2L u+L2

L2

du u

u L

2b

Qb 1 Qb 2 b 1 1 b = du 2 = cos 1 2 u b 2 o b u u b 2 o b

Q 1 b b Q Q = cos cos 1 = 0 = 2b b 2 o 2 o 3 6 o

Physics 9C

Midterm #2 Solutions

March 5, 2013

4. An uncharged, solid cylinder has a length L and is made of conducting material with a uniform resistivity throughout the rod of . It is neatly cut into two pieces, with the cut creating two at, circular surfaces. The two pieces are then pulled directly apart, creating a gap of length d, a distance much smaller than the diameter of the cylinder. A voltage difference is introduced across the outside ends of the two pieces. Derive an expression in terms of L, , and d (as well as any appropriate physical constants) for the time it takes the voltage difference across the two interior (cut) ends to equal one half of the voltage difference across the outside ends. V

d L = L1 + L2 L1 L2 By cutting the cylinder and separating the ends slightly, we have created a parallel-plate capacitor with a capacitance of: A C= o , d where A is the cross-sectional area of the cylinder. So now this looking like a charging RC circuit. We have two resistors (one on each side of the capacitor), but they are in series, so we can simply combine them to an equivalent resistance: L L L R = R1 + R2 = 1 + 2 = A A A The time constant of this circuit is the product of the resistance and the capacitance: A o L oL RC = = d d A The inside potential difference will be half the outside potential difference when half the maximum charge reaches the plates. Putting in Q(t)=Q/2 and solving for time gives the same result derived in lecture in terms of the time constant: L t1/2 = RC ln 2 = o ln 2 d

Physics 9CPart III: Detailed Problems

Midterm #2 Solutions

March 5, 2013

5. An insulator in the shape of a half-spherical shell has has a net charge distributed evenly about its surface. It rests on its at side against an innite conducting plane (with which it cannot exchange charge). Find the ratio of the surface charge density on the insulator to the surface charge density on the conductor evaluated at the center of the circular region of contact. [Hint: Start by computing the electric eld at the origin due to just the hemispherical shell of charge, and then employ the method of images.]

conductorat origin

shell

=?shell of charge

conductor First we need to compute the electric eld at the origin due to the hemisphere of charge. Calling the radius of the hemisphere R and using polar coordinates with the vertical axis as the z-direction, we get the following charge element: dQ = shell dA = shell ( R 2 sin d d ) From symmetry, we know that at the origin there will be no x or y components of electric eld, so only the contribution to the z-component from each charge element is needed. At the origin, the electric eld of a single charge element points directly away from (or toward, if the charge is negative) the element, so the angle the electric eld makes with the z-axis is exactly the polar angle that appears in the line element. The magnitude of the zcomponent of the electric eld at the origin is therefore: dEz = dE cos Plugging in for dE in terms of a single charge element (which are all a distance R from the origin) gives:

[We dont know the sign of the charge on the shell, so for now we will work with absolute values and magnitudes.] Now we only need to integrate over the entire shell to get the total electric eld there (remember, thanks to the symmetry, the total electric eld is entirely in the z-direction once all of the elements are included): 2 /2 1 E = shell sin cos d d = shell ( 2 ) = shell 4 o 4 o 0 4 o 2 0

Physics 9C

Midterm #2 Solutions

March 5, 2013

This is the electric eld magnitude due to the shell at the origin. Now for the method of images... The charge on the conductor will place itself such that it is completely equivalent to place a reected distribution of opposite-signed charge on the other side of the plane, and then use the two charge distributions to determine the eld. The magnitude of the eld at the origin due to the image shell will be the same as that of the actual shell, as the calculation is identical. The direction of the eld of the image shell (because of the opposite sign) will be the same as the direction of the real shell, making the eld twice as strong at the origin: E real +image = 2 E real = shell 2 o We know that the electric eld near a conductor in terms of the charge density on the conductor is: conductor at origin E at origin = o Setting this equal to the electric eld calculated above gives the ratio of the absolute values of the charge densities: conductor at origin shell shell = =2 o 2 o conductor at origin The signs of the charges on the conductor and the shell are of course opposite, so we nally get: shell = 2 conductorat origin

6. A conducting loop of radius 2.0m lies in a plane perpendicular to a uniform magnetic eld of 3.0T. The radius of the loop begins to expand at a rate of 1.50m/s, and an emf is induced in the loop as a result. The loop has a resistance of 7.0 . After 4.0s of expansion: expanding loop

a. Calculate the current (including direction CW or CCW in the diagram) in the loop. The ux through the loop is: B = B d A = BA = B ( r 2 )

Physics 9C

Midterm #2 Solutions

March 5, 2013

The radius is changing with time at a constant rate, so the equation for the radius in terms of the starting radius and the speed (v) with which it is increasing is: r ( t ) = ro + vt Plugging this into the equation for ux gives the ux as a function of time: 2 B ( t ) = B ( ro + vt ) Now apply Faradays Law: